These is yes!!! algebra one or two
Sample space for:
First shooting a basketball. Two options: X or O
Then rolling a die. Six options: 1, 2, 3, 4, 5, or 6
Sample space:
X1
X2
X3
X4
X5
X6
O1
O2
O3
O4
O5
O6
Answer: Options:
A. X1
C. O6
F. X6
Answer:
(2x^3-2x^2-12x) is the required product.
Step-by-step explanation:
The given equation is:
2x(x-3)(x+2)
Solving the above given equation, we get
=(2x^2-6x)(x+2)
which can be written as:
=(2x^3-6x^2+4x^2-12x)
Solving the like terms, we get
=(2x^3-2x^2-12x)
which is the required product of the given expression.
Answer:
the solution is the whole line y = -x + 7
Step-by-step explanation:
both equations are identical
2y = 14 - 2x
=> y = 7 - x
which is exactly the same information as y=-x+7
so, there is no single crossing point. they are exactly covering each other. what "remains" is simply the whole line.
Answer:
Step by Step Solution
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
6*x+11-(21)=0
Step by step solution :
STEP
1
:
Pulling out like terms
1.1 Pull out like factors :
6x - 10 = 2 • (3x - 5)
Equation at the end of step
1
:
STEP
2
:
Equations which are never true:
2.1 Solve : 2 = 0
This equation has no solution.
A a non-zero constant never equals zero.
Solving a Single Variable Equation:
2.2 Solve : 3x-5 = 0
Add 5 to both sides of the equation :
3x = 5
Divide both sides of the equation by 3:
x = 5/3 = 1.667
One solution was found :
x = 5/3 = 1.667